Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : noreply (at) *nospam* example.org (joes)
Groupes : sci.mathDate : 15. Dec 2024, 21:21:57
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <069069bf23698c157ddfd9b62b9b2f632b484c40@i2pn2.org>
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User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Sun, 15 Dec 2024 16:25:55 +0100 schrieb WM:
On 15.12.2024 12:15, joes wrote:
Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:
That pairs the elements of D with the elements of ℕ. Alas, it can
be proved that for every interval [1, n] the deficit of hats
amounts to at least 90 %. And beyond all n, there are no further
hats.
But we aren't dealing with intervals of [1, n] but of the full set.
Those who try to forbid the detailed analysis are dishonest
swindlers and tricksters and not worth to participate in scientific
discussion.
No, we are not forbiding "detailed" analysis
Then deal with all infinitely many intervals [1, n].
??? The bijection is not finite.
Therefore we use all [1, n].
Those are all finite.
The problem is that you can't GET to "beyond all n" in the pairing,
as there are always more n to get to.
If this is impossible, then also Cantor cannot use all n.
Why can't he? The problem is in the space of the full set, not the
finite sub sets.
The intervals [1, n] cover the full set.
Only in the limit.
With and without limit.
Wonrg. There is no natural n that „covers N”.
Yes, there are only 1/10th as many Black Hats as White Hats, but
since that number is Aleph_0/10, which just happens to also equal
Aleph_0, there is no "deficit" in the set of Natual Numbers.
This example proves that aleph_0 is nonsense.
Nope, it proves it is incompatible with finite logic.
There is no other logic.
There is the logic of the infinite.
All logic is finite.
Even that about infinities.
-- Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:It is not guaranteed that n+1 exists for every n.